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NCERT Solutions for Class 8 Maths Exercise 8.3 Class 8 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 8 Maths chapter wise NCERT solution for Maths Book all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

**NCERT solutions for Class 8 Maths Comparing Quantities Download as PDF**

## NCERT Solutions for Class 8 Maths Comparing Quantities

**Class –VIII Mathematics (Ex. 8.3)**

**NCERT SOLUTION**

**1. Calculate the amount and compound interest on:**

**(a) Rs.10,800 for 3 years at **** per annum compounded annually.**

**(b) Rs.18,000 for **** years at 10% per annum compounded annually.**

**(c) Rs.62,500 for **** years at 8% per annum compounded annually.**

**(d) Rs.8,000 for **** years at 9% per annum compounded half yearly. (You could the year by year calculation using S.I. formula to verify).**

**(e) Rs.10,000 for **** years at 8% per annum compounded half yearly.**

**Ans. (a) **Here, Principal (P) = Rs. 10800, Time = 3 years,

Rate of interest (R) =

Amount (A) =

= =

= =

=

= Rs. 15,377.34

Compound Interest (C.I.) = A – P

= Rs. 10800 – Rs. 15377.34 = Rs. 4,577.34

**(b)** Here, Principal (P) = Rs. 18,000, Time = years, Rate of interest (R)

= 10% p.a.

Amount (A) =

= =

= =

= Rs. 21,780

Interest for years on Rs. 21,780 at rate of 10% = = Rs. 1,089

Total amount for years

= Rs. 21,780 + Rs. 1089 = Rs. 22,869

Compound Interest (C.I.) = A – P

= Rs. 22869 – Rs. 18000 = Rs. 4,869

**(c)** Here, Principal (P) = Rs. 62500, Time = = years = 3 years (compounded half yearly)

Rate of interest (R) = 8% = 4% (compounded half yearly)

Amount (A) =

=

=

=

=

= Rs. 70,304

Compound Interest (C.I.) = A – P

= Rs. 70304 – Rs. 62500 = Rs. 7,804

**(d)** Here, Principal (P) = Rs. 8000, Time = 1 years = 2 years(compounded half yearly)

Rate of interest (R) = 9% = (compounded half yearly)

Amount (A) =

=

=

=

=

= Rs. 8,736.20

Compound Interest (C.I.) = A – P

= Rs. 8736.20 – Rs. 8000

= Rs. 736.20

**(e)** Here, Principal (P) = Rs. 10,000, Time = 1 years = 2 years (compounded half yearly)

Rate of interest (R) = 8% = 4% (compounded half yearly)

Amount (A) =

=

=

=

=

= Rs. 10,816

Compound Interest (C.I.) = A – P

= Rs. 10,816 – Rs. 10,000 = Rs. 816

NCERT Solutions for Class 8 Maths Exercise 8.3

**2. Kamala borrowed Rs.26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?**

**(Hint: Find A for 2 years with interest is compounded yearly and then find SI on the 2 ^{nd} year amount for years).**

**Ans. **Here, Principal (P) = Rs. 26,400, Time = 2 years 4 months, Rate of interest (R) = 15% p.a.

Amount for 2 years (A) =

= =

= =

= Rs. 34,914

Interest for 4 months = years at the rate of 15% =

= Rs. 1745.70

Total amount = Rs. 34,914 + Rs. 1,745.70

= Rs. 36,659.70

NCERT Solutions for Class 8 Maths Exercise 8.3

**3. Fabina borrows Rs.12,500 per annum for 3 years at simple interest and Radhaborrows the same amount for the same time period at 10% per annnum, compounded annually. Who pays more interest and by how much?**

**Ans. **Here, Principal (P) = Rs.12,500, Time = 3 years, Rate of interest (R)

= 12% p.a.

Simple Interest for Fabina =

= = Rs. 4,500

Amount for Radha, P = Rs. 12,500, R = 10% and = 3 years

Amount (A) =

= =

= =

= Rs. 16,637.50

C.I. for Radha = A – P

= Rs. 16,637.50 – Rs. 12,500 = Rs. 4,137.50

Here, Fabina pays more interest

= Rs. 4,500 – Rs. 4,137.50 = Rs. 362.50

NCERT Solutions for Class 8 Maths Exercise 8.3

**4. I borrowsRs.12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?**

**Ans. **Here, Principal (P) = Rs.12,000, Time = 2 years, Rate of interest (R) = 6% p.a.

Simple Interest =

= = Rs. 1,440

Had he borrowed this sum at 6% p.a., then

Compound Interest =

=

=

=

=

= Rs. 13,483.20 – Rs. 12,000

= Rs. 1,483.20

Difference in both interests

= Rs. 1,483.20 – Rs. 1,440.00 = Rs. 43.20

NCERT Solutions for Class 8 Maths Exercise 8.3

**5. Vasudevan invested Rs.60,000 at an interest rate of 12% per ann^um compounded half yearly. What amount would he get:**

**(i) after 6 months?**

**(ii) after 1 year?**

**Ans. (i)** Here, Principal (P) = Rs. 60,000,

Time = 6 months = 1 year(compounded half yearly)

Rate of interest (R) = 12% = 6% (compounded half yearly)

Amount (A) =

=

=

=

=

= Rs. 63,600

After 6 months Vasudevan would get amount Rs. 63,600.

**(ii)** Here, Principal (P) = Rs. 60,000,

Time = 1 year = 2 year(compounded half yearly)

Rate of interest (R) = 12% = 6% (compounded half yearly)

Amount (A) =

=

=

=

=

= Rs. 67,416

After 1 year Vasudevan would get amount Rs. 67,416.

NCERT Solutions for Class 8 Maths Exercise 8.3

**6. Arif took a loan of Rs.80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after **** years if the interest is:**

**(i) compounded annually.**

**(ii) compounded half yearly.**

**Ans. (i)** Here, Principal (P) = Rs. 80,000, Time = years, Rate of interest (R) = 10%

Amount for 1 year (A) =

=

=

=

= Rs. 88,000

Interest for year =

= Rs. 4,400

Total amount = Rs. 88,000 + Rs. 4,400 = Rs. 92,400

**(ii)** Here, Principal (P) = Rs.80,000,

Time = year = 3year (compounded half yearly)

Rate of interest (R) = 10% = 5% (compounded half yearly)

Amount (A) =

=

=

=

=

= Rs. 92,610

Difference in amounts

= Rs. 92,610 – Rs. 92,400 = Rs. 210

NCERT Solutions for Class 8 Maths Exercise 8.3

**7. Maria invested Rs.8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find:**

**(i) The amount credited against her name at the end of the second year.**

**(ii) The interest for the third year.**

**Ans. (i)** Here, Principal (P) = Rs. 8000, Rate of Interest (R) = 5%, Time = 2 years

Amount (A) =

=

=

=

=

= Rs. 8,820

**(ii)** Here, Principal (P) = Rs. 8000, Rate of Interest (R) = 5%, Time = 3 years

Amount (A) =

=

=

=

=

= Rs. 9,261

Interest for 3^{rd} year = A – P

= Rs. 9,261 – Rs. 8,820 = Rs. 441

NCERT Solutions for Class 8 Maths Exercise 8.3

**8. Find the amount and the compound interest on Rs.10,000 for **** years at 10% per annum, compounded half yearly.**

**Would this interest be more than the interest he would get if it was compounded annually?**

**Ans. **Here, Principal (P) = Rs. 10000, Rate of Interest (R) = 10% = 5% (compounded half yearly)

Time = years = 3 years (compounded half yearly)

Amount (A) =

=

=

=

=

= Rs. 11,576.25

Compound Interest (C.I.) = A – P

= Rs. 11,576.25 – Rs. 10,000 = Rs. 1,576.25

If it is compounded annually, then

Here, Principal (P) = Rs. 10000, Rate of Interest (R) = 10%, Time = years

Amount (A) for 1 year =

=

=

=

=

= Rs. 11,000

Interest for year = = Rs. 550

Total amount = Rs. 11,000 + Rs. 550

= Rs. 11,550

Now, C.I. = A – P = Rs. 11,550 – Rs. 10,000

= Rs. 1,550

Yes, interest Rs. 1,576.25 is more than Rs. 1,550.

NCERT Solutions for Class 8 Maths Exercise 8.3

**9. Find the amount which Ram will get on Rs.4,096, if he gave it for 18 months at **** per annum, interest being compounded half yearly.**

**Ans. **Here, Principal (P) = Rs. 4096,

Rate of Interest (R) =

= (compounded half yearly)

Time = 18 months = years = 3 years (compounded half yearly)

Amount (A) =

=

=

=

=

= Rs. 4,913

NCERT Solutions for Class 8 Maths Exercise 8.3

**10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum.**

**(i) Find the population in 2001.**

**(ii) What would be its population in 2005?**

**Ans.** **(i) **Here, A_{2003} = Rs. 54,000, R = 5%, = 2 years

Population would be less in 2001 than 2003 in two years.

Here population is increasing.

A_{2003} =

54000 =

54000 =

54000 =

54000 =

48,980 (approx.)

**(ii)** According to question, population is increasing. Therefore population in 2005,

A_{2005} =

=

=

=

=

= 59,535

Hence population in 2005 would be 59,535.

NCERT Solutions for Class 8 Maths Exercise 8.3

**11. In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.**

**Ans. **Here, Principal (P) = 5,06,000, Rate of Interest (R) = 2.5%, Time = 2 hours

After 2 hours, number of bacteria,

Amount (A) =

=

=

=

=

=

= 5,31,616.25

Hence, number of bacteria after two hours are 531616 (approx.).

NCERT Solutions for Class 8 Maths Exercise 8.3

**12. A scooter was bought at Rs.42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.**

**Ans. **Here, Principal (P) = Rs. 42,000, Rate of Interest (R) = 8%, Time = 1 years

Amount (A) =

=

=

=

=

= Rs. 38,640

Hence, the value of scooter after one year is Rs. 38,640.

## NCERT Solutions for Class 8 Maths Exercise 8.3

NCERT Solutions Class 8 Mathematics PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 8 Mathematics includes text book solutions from Class 8 Maths Book . NCERT Solutions for CBSE Class 8 Maths have total 16 chapters. 8 Maths NCERT Solutions in PDF for free Download on our website. Ncert class 8 solutions PDF and Maths ncert class 8 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.

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